[leetcode] 15. 3Sum

题目

Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.

Notice that the solution set must not contain duplicate triplets.

Example 1:

Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]

Example 2:

Input: nums = []
Output: []

Example 3:

Input: nums = [0]
Output: []

Constraints:

• 0 <= nums.length <= 3000
• -105 <= nums[i] <= 105

思路

快慢指针，使用一个指针遍历数组，另外使用快慢指针置于数组的首尾。当三指针之和大于0时，快指针减一；三指针之和小于0时，慢指针进一；等于0时，则保存此时元素。

代码

python版本：

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        res = []
        nums.sort()
        for i in range(len(nums)-2):
            if i > 0 and nums[i] == nums[i-1]:
                continue
            l, r = i+1, len(nums)-1
            while l < r:
                if nums[l]+nums[r]+nums[i] > 0:
                    r -= 1
                elif nums[l]+nums[r] + nums[i] < 0:
                    l += 1
                else:
                    res.append(sorted([nums[l], nums[r], nums[i]]))
                    while l < r and nums[l] == nums[l+1]:
                        l += 1
                    while l < r and nums[r] == nums[r-1]:
                        r -= 1
                    l += 1
                    r -= 1
        return res